Help Installing Speaker Volume Controls ...
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Posted by M100C on December 24, 2005, 10:22 am
Please log in for more thread options I need some help understanding how volume control impedance calculates into the load seen by the amplifier. Here is my scenario, using just one amp channel for illustration purposes: I have an amp rated to drive loads from 6 to 16 ohms ... ideally, I would like to present the amp with an 8 ohm load. I have three, 8 ohm speakers, all with home runs back to the amp. When connected in parallel, they will provide this channel with a load of 2 2/3 ohms (1/Rt = 1/8 + 1/8 + 1/8), which is clearly too low. This is where I start to get a little fuzzy about how the volume control impedance works ... I know need to create another 6 ohms of impedance to bring the total impedance for this channel to just over 8 ohms, but here are my questions: 1) Are I seeing this correctly? If so, I have not been able to identify any volume controls which would create the proper "make-up" impedance. Your suggestions are welcome. 2) Ideally, I'd like to have a volume control with an "off" setting. In this case, I assume that there might be a make-break switch which removes the speaker from the circuit and adds the additional impedance needed to replace the speaker ... is this correct? Thanks for the education, Chris | |||||||||||||
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Posted by on December 25, 2005, 4:33 am
Please log in for more thread options multiplier. If you were to select "4X" on each of your VC's you'd have 1/Rt = 1/32 + 1/32 + 1/32 which is 10-2/3 Ohms and safe for your amplifier. Selecting "2X" on all three would yield a net impedance of 5-1/3 Ohms, too low for the amp. But wait... If you have one room that is larger / needs more volume than the other two. You could set that room for "2X" and the other two rooms for "4X". The result would be 1/Rt = 1/32 + 1/32 + 1/16, which is exactly 8 Ohms. The Russound V/C's I carry have an off position. The side of the autoformer connected to the speakers becomes an open circuit. The other side stays in the circuit, but draws virtually no power from the line side because at that point there's nothing on the load side. Regards, Robert L Bass www.BassBurglarAlarms.com | |||||||||||||
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